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1. La distribucin de velocidades en un ro muy ancho de 2.8 m de profundidadvara desde 0.6 m/s muy cerca del fondo hasta 1.9 m/s en la superficie, deacuerdo con la ecuacin:
45.0
8.23.16.0
y
v
Cuales son los coeficientes y cuya inclusin en las ecuaciones de Bernoulliy Momentum respectivamente permite el anlisis unidimensional correcto.
45.0
8.23.16.0
yv
45.08179.06.0 yv
yB
dAv
A
Qv
yBdyByv
8.2
0
45.0
8179.06.0
yB
dyyBv
8.2
0
45.08179.06.0 8.2
8179.06.08.2
0
45.0
dyyv
Resolviendo la integral por Hewlett Packard la integral se tiene:
smv /5.18.2
2.4 smv /5.1
Av
dAv
3
3
yB
dyyB
3
8.2
0
345.0
5.1
8179.06.0
8.25.1
8179.06.0
3
8.2
0
345.0
dyy
Resolviendo la integral por Hewlett Packard la integral se tiene:
102.18.25.1
415.103
sm/102.1
Av
dAv
2
2
yB
dyyB
2
8.2
0
245.0
5.1
8179.06.0
8.25.1
8179.06.02
8.2
0
245.0
dyy
Resolviendo la integral por Hewlett Packard la integral se tiene:
033.18.25.1
51.62
sm/033.1
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2. En una seccin transversal de un puente, las velocidades medidas de flujofueron medidas con los puntos medios de varias subreas, como se
muestra en la figura. Calcule los valores y para la seccin transversal.
Av
dAv
33
Av
Av
3
3
dAAA dAvAvQ
2
16.5
2
4.18m
mmA
smsmmQ /6.5/16.5
32
1
2
2128
2
6.14.1mm
mmA
smsmmQ /6.15/3.112
32
2
2
3168
2
4.26.1mm
mmA
smsmmQ /24/5.11632
3
2
46.178
2
24.2mm
mmA
smsmmQ /16.28/6.16.17
32
4
2
5148
2
5.12mm
mmA
smsmmQ /4.15/1.114
32
2
2
625.5
2
5.17m
mmA
smsmmQ /25.5/125.5 321
245.70 mAA smdAvAvQ /01.94 3
vAQ smvsm
m
sm
A
Qv /33.1/33.1
45.70
/01.942
3
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45.7033.1
25.51141.16.176.1165.1123.16.513
333333
3
3
Av
Av
097.1097.145.7033.1
94.1813
45.7033.1
25.51141.16.176.1165.1123.16.512
222222
2
2
Av
Av
036.1036.145.7033.1
13.1293
3. If a flow of 1.8 m/s is a carried in flume at velocity of 1.2 m/s compute thedimensions of the cross section if it is:
a) Circular with y = 0.72do.b) Rectangular with the width equal to twice the depth.c) Trapezoidal with b = y and side slope 3 horizontal to 4 vertical.
Q = 1.8 m/s
V = 1.2 m/s
22
3
5.15.1/2.1
/8.1mAm
sm
sm
v
QA
vAQ
a) y = 0.72do.
y do
y = 0.72do do
0
00
0
0
5.0
5.072.0
2cos
2
2
2cos
d
dd
d
dy
44.0
2
cos
5.0
22.0
2
cos0
0
d
d 44.0cos
2
1
79.127
053.4180
21.231
rad
rad
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sendA 20
8
1
2
2
2
04776.2
21.231053.4
85.18m
sen
m
sen
Ad
mdmmdmd 57.157.14776.24776.20
2
0
22
0
0
2
1dP mPmP 18.318.357.1053.4
2
1
mRmm
m
P
AR 47.047.0
18.3
5.1 2
2
0
sendT msenmT 416.1
2
21.23157.1
mDmm
m
T
AD 06.106.1
416.1
5.1 2
b) B = 2y
y
2y
yyAybA 222yA
22 25.1 ym
mymm
y 86.086.02
5.1 2
ybP 2yyP 22 mPmmPyP 46.346.386.044
mRmm
mPAR 433.0433.0
46.35.1
2
mDm
m
y
m
b
m
T
AD 87.0
86.02
5.1
2
5.15.1 222
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c) b = y
4 y
3
b = y
22yybA 22yyyA 222 32 yAyyA
5.132 y mym
my 7071.07071.0
3
5.1 2
75.04
3z
2
12 zybP 2
75.017071.027071.0 mmPmP 475.2
mRmm
m
P
AR 6.06.0
475.2
5.1 2
mDm
m
yzy
m
yzb
m
T
AD 85.0
7071.075.027071.0
5.1
2
5.1
2
5.1 222
4. En un canal que conduce un caudal de 9 m/s; existe una transicin desalida, que sirve para unir una seccin rectangular con una trapezoidal,cuyas dimensiones se muestran en la figura.
Indicar cul es la velocidad en la seccin rectangular, considerar que lasprdidas entre la seccin 1 y 2 son solo por transicin, siendo la frmula para
su clculo:
g
vvH
L
2
3.0
2
2
2
1
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g
vyzH
g
vyz
L
22
2
2
22
2
1
11
g
v
g
vv
g
vym
2
3.12
3.02
2.0
2
2
2
2
2
1
2
1
1
yb
Q
A
QvvAQ
11
3
1
37.2
8.3
/9
yy
smv
2
222yzybA 2
2)3.1(5.13.18.5 mmmA
2
2075.10 mA
smm
sm
A
Qv /893.0
075.10
/92
3
2
2
6.19
)893.0(3.1
6.19
)893.0(37.2
3.06.19
37.2
2.02
2
2
1
2
1
1
yy
ym
34.16.19
8.061.5
3.06.19
61.5
2.02
1
2
1
1
yyym
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34.16.19
8.061.5
3.06.19
61.52.0
2
1
2
1
2
1
1
y
y
yym
212
1
2
1
2
1
134.1
6.19
8.061.53.0
286.02.0 y
y
y
yy
2
1
2
13
1
2
134.1
6.19
24.068.1286.02.0 y
yyy
034.10122.00857.0286.02.0 21
2
1
3
1
2
1 yyyy
02003.0013.11
2
1
3
1 yyy
Resolviendo por Hewlett Packard
y1 = 0.63 m
mDb
yb
T
AD 63.063.0
8.3
63.08.3
1
11
1
1
1
smvsmmm
smv /76.3/76.3
63.08.3
/91
3
1
1
1
1
Dg
vFr
151.1
0638.976.3
1Fr Flujo Supercrtico
mmDm
m
yzb
yzyb
T
AD 04.1038.1
7.9
075.10
3.15.128.5
3.15.13.18.5
2
22
22
2
222
2
2
2
2
2
2
Dg
vFr
128.0
04.18.9
893.02
Fr Flujo Subcrtico
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5. Para unas prdidas en el vertedero de 2Nm/N y un caudal unitario de 10m/s, determinar la elevacin del piso para que ocurra un resalto.
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6. Determine the normal discharges in channels having the following sectionsfor y = 1.5 m; n = 0.014, and S o = 1%0
a) A rectangular section 3.2 m wide.b) A triangular section with a bottom angle equal to 90
c) A trapezoidal section with a bottom of 4 m and side slopes 1 : 1.d) A circular section 72 in diameter.e) A parabolic section having an equation y = 0.3 X2.
y = 1.5 mn = 0.014So = 1%0 = 0.001
a) Wide = 3.2 m
1.5 m
3.2 m
2
yzybA 22 8.48.45.12.3 mAmmmA
ybP 2 mPmmmP 2.62.65.122.3
2
1
0
3
2
SRn
AQ 2
1
0
3
2
SP
A
n
AQ
smQsmm
mmQ /14.9/14.9001.0
2.6
8.4
014.0
8.4332
13
2
22
b) Angle = 90
X
1.5 m
mxmmxm
x5.15.15.145tan
5.145tan
Z = 1
7/28/2019 caudales ejercicios hidraulica
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2
yzA 222 25.225.2)5.1(1 mAmmA
212 zyP mPmmP 91.591.5115.12 2
2
1
0
3
2
SRn
AQ 2
1
0
3
2
SP
A
n
AQ
smQsmm
mmQ /58.4/58.4001.0
91.5
25.2
014.0
25.2332
13
2
22
c) b = 4 m
1 : 1 25.1125.1
VHz
1 1.5 m1.25
4 m
2
zyybA 222 81.881.8)5.1(25.15.14 mAmmmmA
212 zybP
mPmmmP 8.88.8)25.1(15.1242
2
1
0
3
2
SRn
AQ 2
1
0
3
2
SP
A
n
AQ
smQsmm
mmQ /91.19/91.19001.0
8.8
81.8
014.0
81.833
2
13
2
22
d) d0 = 72 in
inyinm
inmyinin
05.5905.590254.015.1
)()(
23 in59 in 72 in
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63.0cos236
23
2cos 1
in
in 58.10063.02
58.100
radrad
rad 53.4180
42.259
sendA 20
8
1
22 93.357042.25953.4728
1inseninA
22
2
2
2
)(8.248.24
)12(
193.3570 FtAFt
in
FtinA
Ft
0
2
1dP inPininP 08.16308.1637253.4
2
1
FtAFtni
FtniP
Ft59.1359.13
12
108.163
)(
2
1
0
3
2
49.1 SRn
AQ 2
1
0
3
2
49.1 SP
A
n
AQ
sFtQsFtFt
FtFtQ /64.124/64.124001.0
59.13
59.13
014.0
59.1349.1 332
13
2
22
e) y = 0.3X2
37.137.18.0
5.1 Txx
yTA3
2
xA 3.037.13
222 375.0001.137.13.037.1
3
2mAmA
X > 1
22 1
11
2xxLn
xx
TP
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22 )37.1(137.1
37.1
1)37.1(137.15.0 LnP
mP 718.1
2
1
0
3
2
SRn
AQ 2
1
0
3
2
SP
A
n
AQ
smQsmm
mmQ /301.0/301.0001.0
718.1
375.0
014.0
375.0332
13
2
22
7. Using the Manning formula, determine the normal depths in channels havingthe following section when Q = 2 m/s, n = 0.017, S0 = 2%o.
a) A rectangular section 4 m wide.b) A triangular section with a bottom angle equal to 90c) A trapezoidal section with a bottom width of 3.8 m and side slopes 2:1.d) A circular section 1.5 m diameter.e) A parabolic section having an equation y = 0.8 X2.
Q = 2 m/sn = 0.017So = 2%0 = 0.002
a) Wide = 4m
1.5 m
3.2 m
ybA
ybP 2
2
1
0
3
2
SRn
AQ
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2
13
2
3 )002.0(2017.0
/2
yb
ybybsm
044.024
4017.0
4/2
3
2
3
yyysm
3
2
3
)2(
2
)044.0(4
017.0/2
y
yy
m
sm 32
)2(
2193.0
y
yy
Resolviendo por Hewlett Packard:
y = 0.4 m
b) Angle = 90
Z = 1
2
yzA
2
12 zyP
2
1
0
3
2
SRn
AQ 2
1
0
3
2
SP
A
n
AQ
2
13
2
2
3
)002.0(12017.0
/2
zy
yzyzsm
2
13
2
2
2
3 )002.0(112
1
017.0
1/2
y
yysm
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2
13
2
2
2
3 )002.0(112
1
017.0
1/2
y
yysm
044.083.2
82.58/23
2
2
23
y
yysm
3
2
2
3
044.05.082.58
/2yy
sm
myyy 175.154.154.1 8 338
c) b = 3.8 m
2 : 12
1
2
V
Hz
1 y2
3.8 m2zyybA
212 zybP
2
1
0
3
2
SRn
AQ 2
1
0
3
2
SP
A
n
AQ
2
13
2
2
22
3
)002.0(12017.0
/2
zyb
yzybyzybsm
044.02128.3
28.3
017.0
28.3/2
3
2
2
22
3
y
yyyysm
044.02128.3
28.3)65.11753.223(/2
3
2
2
2
23
y
yyyysm
Resolviendo por Hewlett Packard:y = 0.372 m
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d0 = 1.5 m
2
0d
y
y d0 = 1.5 m
sendA 20
8
1
01
4
1d
senR
2
1
0
32
SRnAQ 2
1
0
3
2
SPA
nAQ
21
3
2
2
3 002.05.1125.0017.0
5.1125.0/2
sensensm
044.05.1
180
125.0017.0
5.1180
125.0
/2
3
2
2
3
sensen
sm
Resolviendo por Hewlett Packard:
26.13226.13274.227360360
74.227
75.0
75.0
2cos
y75.075.0
2cos
y
mymy 053.1053.175.075.02
26.132cos
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d) y = 0.8x
yTA 3
2
22
2
832
yTyTR
Si 10 xx
yT
T
yx
44
xTxx
xT
2.32.38.04
2
2
1
0
3
2
SRn
AQ 2
1
0
32
SP
A
n
AQ
2
13
2
22
2
3 )002.0(83
2
017.0
3
2
/2
yT
yTyT
sm
044.08.082.33
8.02.32017.0
8.02.3
3
2
/23
2
222
22
2
3
xxxx
xx
sm
044.012.56.9
38.16
017.0
706.1/2
3
2
42
43
3
xx
xxsm
3
2
2
2
3
3
12.56.9
38.16
044.035.100
/2
x
xx
sm
3
2
2
2
3
12.56.9
38.1645.0
x
xx
Resolviendo por Hewlett Packard:x = 0645
mymyxy 516.0516.0)645.0(8.08.0 22
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8. Un canal trapezoidal revestido de concreto (n = 0.014) de ancho de solera0.5 m, talud 3H:4V y trazado con una pendiente del 1%o, conduce un caudalde 700 Lit/s, en cierto tramo del canal tiene que atravesar una carretera, conuna alcantarilla (n=0.014) de 1.5 m de dimetro.
Para unir la seccin trapezoidal con la circular se construye una transicin conla misma pendiente del canal y de 10 m de longitud. Si las prdidas en latransicin son despreciables, indicar:
a) Cul es la velocidad a la entrada de la alcantarilla?b) Cul es la pendiente con la que se debe trazar la alcantarilla para
conseguir que se establezca el flujo uniforme?
n = 0.014b = 0.5 m3H:4VS0 = 1%o = 0.001Q = 700 Lit /s = 0.7 m/s
d0 = 1.5 m
a)
2
1
0
3
2
SP
A
n
AQ
2
13
2
2 )001.0(12
)(
014.0
)(
zybyyzbyyzb
Q
316.075.0125.0
)75.05.0(
014.0
)75.05.0(/7.0
3
2
2
3
y
yymyymsm
Resolviendo por Hewlett Packard:
y = 0.67 m
0
3
2
1S
P
A
n
v
001.012
)(
014.0
13
2
2
zyb
yyzbv
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smv /032.1001.075.0167.025.0
67.0)67.075.05.0(
014.0
1 32
2
smv /032.1b) d0 = 1.5 m
2
0d
y
y d0 = 1.5 m
575.0
575.0
2cos
y
575.0
575.067.0
2cos
165.0cos2
1
98.160
360 02.19998.160360
radrad
rad 473.3180
02.199
2
1
0
3
2
SRn
A
Q
21
0
3
2
2
3 5.1125.0017.0
5.1125.0/7.0 S
sensensm
21
0
3
2
2
3
5.1473.3
02.199125.0
017.0
5.102.199473.3125.0/7.0 S
sensensm
21
0
3 273.0857.44/7.0 Ssm
oSSSms
%2.30032.00032.00571.0273.0857.44
7.00
2
0
2
1
0
/3
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9. A channel with the cross section shown in the following figure has aMannings coefficient of 0.040 from station (0) to station (3) and 0.054 fromstation (3) to station (8).
The flow through the channel is 13 m/s, and the water surface is 1.7 m high.
Find the following:
a) Weighted Mannings Coefficientb) Slope of the channel.c) Top channel.d) Wetted Perimeter.e) Flow regime (supercritical or subcritical)
a)
32
1
5.1
Pt
pnn
n
iii
e
P1 =X
P1 0.2
mxx
4.01
5.02.0
mPP 447.0447.04.02.01
22
1
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P2 = 1 m
P3 =
1.5 P3
1
mPmP 8.18.115.13
22
3
P4 = 2 m
P5 =
P5 1
1
mPmP 41.141.111 522
5 P6 = 1 m
P7 =x
0.7 m P7
mxx 7.01
1
7.0
mPmP 99.099.07.07.07
22
7
PTotal = 6.937m
5.15.15.15.15.15.15.1
937.6
054.01054.041.1054.02054.08.104.0104.0447.004.0
en
0567.0937.6
09373.0 32
en
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b)
2
121*2 mA 2
275.0
2
5.1*1mA 2
35.0
2
1*1mA
2
48.27.0*4 mA 2
54.02*2.0 mA 2
604.0
24.0*2.0 mA
2
7245.0
2
7.0*7.0mA
ATotal = 6.735 m
2
1
0
3
2
SP
A
n
AQ
2
1
0
3
2
2
22
3
937.6
735.6
0567.0
735.6/13 S
m
mmsm
%21.10121.00121.011.078.118*98.0
/130
2
0
3
2
1
0 SS
smS
c) mTT 1.77.04.06
d) mPP 647.899.0141.128.11447.0 e)
Dg
vFr
mm
m
T
AD 948.0
1.7
735.6 2
smvsm
m
sm
A
QvvAQ /04.2/04.2
375.6
/132
3
66.0948.0*8.9
/04.2
smFr
166.0Fr Flujo Subcrtico
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10. A stream with the cross section shown in the previous figure has a flowrate of 5 m/s. The stream has a longitudinal slope of 0.002m/m and anatural stony bottom (n = 0.05, stations 0 to 8)
a) Using Mannings equation, what is the water surface elevation of thestream?
b) What is the maximum capacity of the channel?c) How would the capacity of the channel be affected if you were to
pave the center of the channel (n = 0.013) between stations 3 and 5?
Q = 5 m/sn = 0.05S0 = 0.002
a)
Suponiendo un y = 1.5 se tiene:
mP 8.15.1122
1
mP 22
mP 41.111 22
3
mP 14
mP 7071.05.05.0 22
5
mmmmmmPTotal 924.67071.0141.128.1 2
121*2 mmmA
2
275.0
2
5.1*1m
mmA
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2
35.0
2
1*1m
mmA
2
425.0*4 mmmA
2
5125.0
25.0*5.0 mmmA
222222 375.5125.025.075.02 mmmmmmATotal
2
1
0
3
2
SP
A
n
AQ
smQ /0607.4)002.0(
924.6
375.5
05.0
375.52
13
2
Suponiendo un y = 1.7 se tiene:
mP 8.15.11 221
mP 2
2
mP 41.111 223
mP 1
4
mP 99.07.07.0 225
mP 1
6
mP 447.02.04.0 227
mmmmmmmmP
Total654.8447.0199.0141.128.1
2
121*2 mmmA
2
275.0
2
5.1*1m
mmA
2
35.0
2
1*1m
mmA
2
48.27.0*4 mmmA
2
5
245.02
7.0*7.0m
mmA
2
64.02.0*2 mmmA
2
704.0
2
2.0*4.0m
mmA
22222222 735.604.04.0245.08.25.075.02 mmmmmmmmATotal
7/28/2019 caudales ejercicios hidraulica
24/27
2
1
0
3
2
SP
A
n
AQ
smQ /09.5)002.0(
654.8
735.6
05.0
735.62
13
2
Interpolando los dos datos se tiene:y Q
0607.45.1
09.57.1
mysmQ 68.1/5 3
Suponiendo un y = 1.68 se tiene:
mP 8.15.1122
1
mP 22
mP 41.111 22
3
mP 14
mP 96.068.068.0 22
5
mP 16
mP 40245.018.036.0 22
7
mmmmmmmmPTotal
572.840245.0196.0141.128.1
2
121*2 mmmA
22 75.0
2
5.1*1m
mmA
2
35.0
2
1*1m
mmA
2
472.268.0*4 mmmA
2
52312.0
2
68.0*68.0m
mmA
2
636.018.0*2 mmmA
2
70324.0
218.*376.0 mmmA
22222222 5936.60324.036.02312.072.25.075.02 mmmmmmmmATotal
2
1
0
3
2
SP
A
n
AQ
smQ /947.4)002.0(
57245.8
5936.6
05.0
5936.62
13
2
7/28/2019 caudales ejercicios hidraulica
25/27
Interpolando Nuevamente se tiene:y Q
9475.468.1
09.57.1
mysmQ 688.1/5 3
Suponiendo un y = 1.688m se tiene:
mP 8.15.1122
1
mP 22
mP 41.111
22
3
mP 14
mP 97.0688.0688.022
5 mP 1
6
mP 42038.0188.0376.0 227
mmmmmmmmP
Total61.842038.0197.0141.128.1
2
121*2 mmmA
2
275.0
2
5.1*1m
mmA
2
35.0
21*1 mmmA
2
4752.2688.0*4 mmmA
2
52366.0
2
688.0*688.0m
mmA
7/28/2019 caudales ejercicios hidraulica
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2
6376.0188.0*2 mmmA
2
70353.0
2
188.0*376.0m
mmA
22222222
65.60353.0376.02366.0752.25.075.02 mmmmmmmmATotal
2
1
0
3
2
SP
A
n
AQ
smsmQ /5/006.5)002.0(61.8
65.6
05.0
65.632
13
2
OK
y = 1.688m
b) y = 2 m se tiene:
mP 8.15.11 221
mP 2
2
mP 41.111 223
mP 1
4
mP 41.111 225
mP 1
6
mP 118.15.05.0 227
mmmmmmmmP
Total738.8118.1141.1141.128.1
2
121*2 mmmA
2
275.0
2
5.1*1m
mmA
2
35.0
2
1*1m
mmA
2
441*4 mmmA
2
55.0
2
1*1m
mmA
2
615.0*2 mmmA
2
725.0
2
5.0*1m
mmA
7/28/2019 caudales ejercicios hidraulica
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22222222 925.015.045.075.02 mmmmmmmmATotal
2
1
0
3
2
SP
A
n
AQ
smQ /21.8)002.0(
738.8
9
05.0
92
13
2
c)
32
1
5.1
Pt
pnn
n
iii
e
3
2
5.15.15.15.15.15.15.1
61.8
97.005.0105.041.105.02013.08.105.0105.04204.005.0
en
ne = 0.0089
2
1
0
3
2
SP
A
n
AQ
smQ /13.28)002.0(
61.8
65.6
05.0
65.6 2132
smQ /13.28
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